Question

Simplify \( AB + A\bar{C} + \overline{A}BC(AB+C) \)

• A. \( \bar{A} + \bar{B} + AC \)
• B. \( AB + \overline{A}BC \)
• C. \( ABC \)
• D. 1

Hint:

Use Boolean algebra laws: \(X\overline{X}=0\), \(XX=X\), \(X+XY=X\), \(X(X+Y)=X\), DeMorgan's.
Karnaugh maps (K-maps) are useful for simplifying expressions with 2-4 variables.
If an MCQ answer seems incorrect based on your derivation, double-check your work, then consider if there's a typo in the question or options.

Answer 1

The Correct Option is D

Let the expression be Y. \( Y = AB + A\bar{C} + \overline{A}BC(AB+C) \) First, expand the last term: \( \overline{A}BC(AB+C) = \overline{A}BC \cdot AB + \overline{A}BC \cdot C \) Since \(A\overline{A}=0\) and \(CC=C\): \( \overline{A}BC \cdot AB = A\overline{A}BBC = 0 \cdot B \cdot B \cdot C = 0 \). \( \overline{A}BC \cdot C = \overline{A}BCC = \overline{A}BC \). So, the last term simplifies to \(\overline{A}BC\). The expression becomes: \( Y = AB + A\bar{C} + \overline{A}BC \) We can use consensus theorem or Karnaugh map. Let's use Boolean algebra: \( Y = AB + A\bar{C} + \overline{A}BC \) Consider \(A\bar{C} + \overline{A}BC\). This is not directly simplifying. Let's try to factor: \( Y = A(B + \bar{C}) + \overline{A}BC \) This is not obviously leading to 1. Let's re-check the expansion of the last term: \(\overline{A}BC(AB+C)\) If the term was \(\overline{A}BC + (AB+C)\) or similar, it would be different. The problem states \(\overline{A}BC(AB+C)\). So \( \overline{A}BC(AB) + \overline{A}BC(C) \) \( = (\overline{A}A)B^2C + \overline{A}BC^2 \) \( = 0 \cdot B \cdot C + \overline{A}BC = 0 + \overline{A}BC = \overline{A}BC \). So \( Y = AB + A\bar{C} + \overline{A}BC \). Is there a typo in the question or options, if the answer is 1? Let's test the expression \( AB + A\bar{C} + \overline{A}BC \) for some values: A=1, B=1, C=0: \(1 \cdot 1 + 1 \cdot 1 + 0 \cdot 1 \cdot 0 = 1+1+0 = 1\). (Boolean sum) A=0, B=1, C=1: \(0 \cdot 1 + 0 \cdot 0 + 1 \cdot 1 \cdot 1 = 0+0+1 = 1\). A=1, B=0, C=0: \(1 \cdot 0 + 1 \cdot 1 + 0 \cdot 0 \cdot 0 = 0+1+0 = 1\). A=1, B=1, C=1: \(1 \cdot 1 + 1 \cdot 0 + 0 \cdot 1 \cdot 1 = 1+0+0 = 1\). A=0, B=0, C=0: \(0 \cdot 0 + 0 \cdot 1 + 1 \cdot 0 \cdot 0 = 0+0+0 = 0\). Since we found a case (A=0,B=0,C=0) where Y=0, the expression cannot simplify to 1. Therefore, the marked answer (d) 1 is incorrect for the given expression. Let's check if any standard simplification applies to \(AB + A\bar{C} + \overline{A}BC\). This expression is the sum of products: \(m_7, m_6\) (from AB); \(m_5, m_4\) (from A\(\bar{C}\)); \(m_3\) (from \(\overline{A}BC\)). (Assuming A is MSB: ABC) m7 = 111, m6 = 110 m5 = 101, m4 = 100 m3 = 011 So \(Y = \sum m(3,4,5,6,7)\). Using K-map for 3 variables A, B, C: \begin{tabular}{c|cccc} \textit{A\textbackslash BC} & 00 & 01 & 11 & 10
\hline 0 & 0 & 0 & 1 & 0
1 & 1 & 1 & 1 & 1
\end{tabular} Minterms: m0, m1, m3, m2 for row A=0. m4, m5, m7, m6 for row A=1. \(Y = m_3 + m_4 + m_5 + m_6 + m_7\) \(m_3 = \overline{A}BC\) \(m_4 = A\overline{B}\overline{C}\) \(m_5 = A\overline{B}C\) \(m_6 = AB\overline{C}\) \(m_7 = ABC\) The terms in the expression were: \(AB = ABC + AB\overline{C} = m_7 + m_6\) \(A\overline{C} = A\overline{B}\overline{C} + AB\overline{C} = m_4 + m_6\) \(\overline{A}BC = m_3\) So \(Y = (m_7+m_6) + (m_4+m_6) + m_3 = m_3 + m_4 + m_6 + m_7\). (Note: \(A\bar{C}\) covers \(A\bar{B}\bar{C}\) and \(AB\bar{C}\)). Ah, \(A\bar{C}\) has B as don't care in that term for the K-map representation. \(AB\) covers cells (A=1, B=1, C=0) and (A=1, B=1, C=1) \(A\bar{C}\) covers cells (A=1, B=0, C=0) and (A=1, B=1, C=0) \(\overline{A}BC\) covers cell (A=0, B=1, C=1) The K-map is: BC A 00 01 11 10 -------------- 0 | 0 | 0 | 1 | 0 | (\(\overline{A}BC\) for 011) -------------- 1 | 1 | 0 | 1 | 1 | (AB for 110, 111; A\(\overline{C}\) for 100, 110) -------------- So the 1s are at: \(\overline{A}BC\) (011), \(A\overline{B}\overline{C}\) (100), \(AB\overline{C}\) (110), \(ABC\) (111). Correct sum of minterms: \(Y = \sum m(3, 4, 6, 7)\). The expression was \(Y = AB + A\bar{C} + \overline{A}BC\). Grouping on K-map: Pair of \(m_6, m_7 \rightarrow AB\). Pair of \(m_4, m_6 \rightarrow A\overline{C}\). (This is already how it was, the K-map shows these are prime implicants). \(m_3 = \overline{A}BC\). So the simplified expression is indeed \(AB + A\bar{C} + \overline{A}BC\). It doesn't simplify further to any of the other options easily, and certainly not to 1. There must be a typo in the original expression if the answer is 1. For example, if it was \(AB + A\bar{C} + BC\). \(AB+BC = B(A+C)\). So \(A\bar{C} + B(A+C)\). Not 1. If it was \(A+ \overline{A} = 1\). If the question was something like \(A + \overline{A}B + \overline{A}\overline{B}C\). This is \(A + \overline{A}(B+\overline{B}C) = A + \overline{A}(B+C)\) using \(X+\overline{X}Y = X+Y\). Then \(A+B+C\). Not 1. Given the marked answer is (d) 1, and my simplification does not yield 1, there's a high probability of an error in the question statement or the provided correct answer. I will state the simplified form I found. \[ \boxed{\parbox{0.9\textwidth}{\centering The expression \(AB + A\bar{C} + \overline{A}BC\) does not simplify to 1. Its simplified SOP form is \(AB + A\overline{C} + \overline{A}BC\). There might be an error in the question or options.}} \]